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प्रश्न
A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is ______.
पर्याय
`1/2`
1
2
4
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उत्तर
A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is 1.
Explanation:
Consider two amperian loops of radius `a/2` and 2a. Applying Ampere's circuital law for these loops.
We get `ointB.dL = μ_0I_{"enclosed"}`
For the smaller loop,
`B_1 xx 2pi a/2 = mu_0 xx pia^2 1 xx (a/2)^2`
`mu_0I xx 1/4 = (mu_0I)/4`
`B_1 = (mu_0I)/(4pia)`
`B_2 xx 2pi(2a) = mu_0I`
`B_2 = (mu_0I)/(4pia)`
∴ B1/B2 = `(mu_0I)/(4pia) xx (4pia)/(mu_0I) = 1`
संबंधित प्रश्न
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|
Consider the experimental set-up shown in the figure. This jumping ring experiment is an outstanding demonstration of some simple laws of Physics. A conducting non-magnetic ring is placed over the vertical core of a solenoid. When current is passed through the solenoid, the ring is thrown off. |
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Substituting \[B=\frac{\mu_0 I}{2\pi r}\] into \[\oint\vec{B}\cdot d\vec{l}=B(2\pi r)\] for a straight wire yields:
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